[面试SQL试题]查询出在半个小时内通话多次（2次及以上）的用户id

这是最近一次面试，遇到的试题之一。后来开机对该题做了复现，记录如下。

有一张记录了用户通话时间的表records，共三个字段：call_day，call_time, user_id，分别为通话日期，通话时间，用户id。
写出SQL语句，查询出在半个小时内通话多次（2次及以上）的用户id。

字段	类型
call_day （通话日期）	date
call_time （通话时间）	time
user_id（用户id）	int

1、在MySQL创建表records

CREATE TABLE records
(
call_day 	date 	NOT NULL,
call_time 	time 	NOT NULL,
user_id 	int 	NOT NULL
);

2、插入数据

INSERT INTO records VALUES('2020-2-13', '11:24:23', 2234);
INSERT INTO records VALUES('2020-2-13', '13:26:23', 2235);
INSERT INTO records VALUES('2020-2-13', '11:14:24', 2263);
INSERT INTO records VALUES('2020-2-13', '11:44:24', 2263);
INSERT INTO records VALUES('2020-2-13', '11:25:23', 2263);
INSERT INTO records VALUES('2020-2-14', '12:32:23', 2236);
INSERT INTO records VALUES('2020-2-14', '10:42:23', 2237);
INSERT INTO records VALUES('2020-2-14', '11:52:23', 2293);
INSERT INTO records VALUES('2020-2-15', '12:26:23', 2283);
INSERT INTO records VALUES('2020-2-16', '14:32:23', 2633);
INSERT INTO records VALUES('2020-2-17', '14:23:23', 2243);
INSERT INTO records VALUES('2020-2-17', '14:27:23', 2263);
INSERT INTO records VALUES('2020-2-17', '13:29:23', 2633);
INSERT INTO records VALUES('2020-2-17', '17:24:23', 2833);
INSERT INTO records VALUES('2020-2-18', '10:25:23', 2563);
INSERT INTO records VALUES('2020-2-18', '18:42:23', 2893);
INSERT INTO records VALUES('2020-2-18', '18:44:24', 2893);
INSERT INTO records VALUES('2020-2-20', '10:26:23', 2233);
INSERT INTO records VALUES('2020-2-20', '10:12:45', 2233);
INSERT INTO records VALUES('2020-2-20', '10:43:51', 2233);
INSERT INTO records VALUES('2020-2-21', '11:22:23', 2232);
INSERT INTO records VALUES('2020-2-21', '11:32:23', 2232);
INSERT INTO records VALUES('2020-2-21', '11:12:23', 2232);

3、自连接

SELECT 	a.call_day, 
		a.user_id, 
		a.call_time as t1, 
		b.call_time as t2
FROM records a LEFT JOIN records b
ON a.call_day = b.call_day AND a.user_id = b.user_id
WHERE a.call_time > b.call_time;

4、计算时间差

SELECT	a.call_day, 
		a.user_id, 
		a.call_time as t1, 
		b.call_time as t2,
		(a.call_time - b.call_time) AS t1_t2
FROM records a LEFT JOIN records b
ON a.call_day = b.call_day AND a.user_id = b.user_id
WHERE a.call_time > b.call_time;

5、查询半小时内的记录

SELECT	a.call_day, 
		a.user_id, 
		a.call_time as t1, 
		b.call_time as t2,
		(a.call_time - b.call_time) AS t1_t2
FROM records a LEFT JOIN records b
ON a.call_day = b.call_day AND a.user_id = b.user_id
WHERE (a.call_time - b.call_time) > 0 AND (a.call_time - b.call_time) < 3000;